What is the conditional probabilty of event A and B occurring given that exactly three of the events occur?

OK, so I guess I'm supposed to use Bayes' theorem or something like that, right? I know that in the denomenator the number will be (0,4*0,8*0,7*0,8) + (0,4*0,8*0,3*0,2) + (0,4*0,2*0,7*0,2) + (0,6*0,8*0,7*0,2) since these are the probabilities of all possible "three events occur" situations. However, I am undcertain as to what I should put in the numerator. I guess it would be something like P(three events l A ∩B)*P(A ∩B)). But I really don't know how to calculate this.

Or am I overcomplicating the whole situation?

Sep 14th 2009, 06:52 AM

krje1980

Hm, I think perhaps I overcomplicated the situation.

Instead of Bayes' theorem I just use the formula for conditional probability which says that I am supposed to find:

P((A∩B)∩(three events occur)) / P(three events occur))

Thus, I get (0,4*0,8*(three events)) / (three events)

(Note: this is because the events are classified as independent, in which case P(A ∩B) = P(A)*P(B)

The probabilities related to three events cancel each other out, and thus I am left with 0,4*0,8 = 0,32.

Is this correct?

Sep 14th 2009, 07:54 AM

Plato

Quote:

Originally Posted by krje1980

Consider 4 indpendent events with the following probabilty of each event occurring: P(A) = 0,4 P(B) = 0,8 P(C) = 0,7 P(D) = 0,2
What is the conditional probabilty of event A and B occurring given that exactly three of the events occur? I know that in the denomenator the number will be (0,4*0,8*0,7*0,8) + (0,4*0,8*0,3*0,2) + (0,4*0,2*0,7*0,2) + (0,6*0,8*0,7*0,2) since these are the probabilities of all possible "three events occur" situations. However, I am undcertain as to what I should put in the numerator.

Would the numerator be (0,4*0,8*0,7*0,8) + (0,4*0,8*0,3*0,2)?
WHY? or WHY NOT?

Sep 14th 2009, 10:58 AM

krje1980

Hi.

You know, the funny thing is that your solution is what I attempted at first, but then when I noticed the word "independent" I started questioning the reasoning. But I guess it makes sense in that the numerator should actually be P((A∩3 events)∩(B∩3 events)). This is the same as you and the same as what I originally thought.

This is why I don't like probabilities. They mess with my mind! Give me calculus any day :) *